You have found the following ages (in years) of 5 porcupines. Those porcupines were randomly selected from the 41 porcupines at your local zoo: $ 6,\enspace 10,\enspace 19,\enspace 6,\enspace 12$ Based on your sample, what is the average age of the porcupines? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 41 porcupines, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{6 + 10 + 19 + 6 + 12}{{5}} = {10.6\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {21.16} + {0.36} + {70.56} + {21.16} + {1.96}} {{5 - 1}} $ {s^2} = \dfrac{{115.2}}{{4}} = {28.8\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{28.8\text{ years}^2}} = {5.4\text{ years}} $ We can estimate that the average porcupine at the zoo is 10.6 years old. There is also a standard deviation of 5.4 years.